Use the law of corresponding states and Figure 16.10 to estimate the molar volum

Question

Use the law of corresponding states and Figure 16.10 to estimate the molar volume of ethaneat T = 366K and P = 97.4bar.At the same corresponding state, what is the temperature and pressure for ethene? (Critical parameters are in Table 16.5)

TR 2.00 TR 1.50 0.8 R 1.20 0.6 0.4 e Nitrogen -Butane Methane Isopentane 4 Ethane Heptane Ethylene Carbon dioxide 0.2 * Propane Water R 1.00 0.0 FIGURE 16.10 An illustration of the law of corresponding states. The compressibility factor, Z, is plotted against the reduced pressure, P of each of the 10 indicated gases. Each curve represents a given reduced temperature. Note that for a given reduced temperature, all 10 gases fall on the same curve because reduced quantities are used.

Explanation / Answer

Law of corresponding states

At the same reduced temperature and reduced pressure, the compressibility factor also remains the same for real gases.

reduced temperature

Tr = T/Tc

Reduced pressure

Pr = P/Pc

Reduced molar Volume

Vr = V/Vc

Part a

T = temperature = 366 K

P = pressure = 97.4 bar

V = molar volume.

c subscript represents critical point

For ethane

Tc = 305.34 K

Pc = 48.714 bar

Tr = 366/305.34 = 1.198

Pr = 97.4/48.714 = 1.999

From the graph

Compressibility factor z = 0.58

z = PV/RT

Molar volume

V = zRT/P

= 0.58 x 8.314*10^-2 L-bar/mol·K x 366 K / 97.4 bar

= 0.1812 L/mol

Part b

For ethene

Tc = 282.35 K

Pc = 50.422 bar

At the same reduced properties

T = Tr x Tc

= 1.198 x 282.35

= 338.255 K

P = Pr x Pc = 1.999 x 50.422

= 100.79 bar

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